What are the solutions to the equation (2x+1)(x+5)=-3 ?

The solutions to the equation (2x+1)(x+5)=-3 are x=(-11+sqrt(57))/4 ,x=(-11-sqrt(57))/4

Find the zeros of (2x+1)(x+5)=-3

The zeros of (2x+1)(x+5)=-3 are x=(-11+sqrt(57))/4 ,x=(-11-sqrt(57))/4

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(2 x + 1) (x + 5) = - 3

x = \frac{- 11 + 57}{4}, x = \frac{- 11 - 57}{4}

Decimal Notation

x = - 0.86254 \dots, x = - 4.63745 \dots

Solution steps

(2 x + 1) (x + 5) = - 3

Expand

(2 x + 1) (x + 5) : 2 x^{2} + 11 x + 5

2 x^{2} + 11 x + 5 = - 3

Move

3

to the left side

2 x^{2} + 11 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

1 1^{2} - 4 \cdot 2 \cdot 8 = 57

x_{1, 2} = \frac{- 11 \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 11 + 57}{2 \cdot 2}, x_{2} = \frac{- 11 - 57}{2 \cdot 2}

x = \frac{- 11 + 57}{2 \cdot 2} : \frac{- 11 + 57}{4}

x = \frac{- 11 - 57}{2 \cdot 2} : \frac{- 11 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{- 11 + 57}{4}, x = \frac{- 11 - 57}{4}

\frac{3}{11} \cdot 7

f = 0^{2}

t^{2} - 1

9 x^{2} - 36 x = 0

What are the solutions to the equation (2x+1)(x+5)=-3 ?
The solutions to the equation (2x+1)(x+5)=-3 are x=(-11+sqrt(57))/4 ,x=(-11-sqrt(57))/4
Find the zeros of (2x+1)(x+5)=-3
The zeros of (2x+1)(x+5)=-3 are x=(-11+sqrt(57))/4 ,x=(-11-sqrt(57))/4